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Motion In Straight Line

Question
CBSEENPH11017682

A rocket is fired vertically with a speed of v km/s from the surface of the earth. Find the maximum height attained by the rocket before it returns to the earth. Take GM =gR2 at the surface of the earth.

Solution
Let a rocket of mass m be projected from the surface of the earth with velocity v less than escape velocity of satellite.
Let h be the maximum height attained by satellite before it returns to the surface of the earth.
Using law of conservation of the energy,
Total energy at surface of earth = total energy at height h
i.e.         1 half mv squared minus GMm over straight R equals 0 minus fraction numerator GMm over denominator straight R plus straight h end fraction 
rightwards double arrow           space space 1 half mv squared minus mgR equals 0 minus fraction numerator mgR squared over denominator straight R plus straight h end fraction 
                   space space space space fraction numerator straight v squared minus 2 gR over denominator 2 end fraction equals negative fraction numerator gR squared over denominator straight R plus straight h end fraction 
rightwards double arrow                      space space space space straight R plus straight h equals fraction numerator 2 gR squared over denominator left parenthesis 2 gR minus straight v squared right parenthesis end fraction 
∴        straight h equals fraction numerator 2 gR squared over denominator left parenthesis 2 gR minus straight v squared right parenthesis end fraction minus straight R equals fraction numerator Rv squared over denominator 2 gR minus straight v squared end fraction
'h' is the maximum height attained by the rocket before it returns to the Earth.